- #1
magnifik
- 360
- 0
this is the problem:
x2y' = (2y2 - x2)
here's what i have done so far:
dy = (2y2/x2 - 1)dx
(2y2/x2 - 1)dx - dy = 0
i used the substitution y = xv then found an integrating factor and got
dx/x - dv/(2v2 - 2v - 1) = 0
but i am stuck at this point..
i know ln(x) + C is the first part, but is there an easy way to find the second part? or am i totally off to begin with?
this is the solution:
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP50419e2227898a2f2f0000026hh297a4ebd6b3h?MSPStoreType=image/gif&s=33&w=154&h=47
i am at a loss for how they got to this point
x2y' = (2y2 - x2)
here's what i have done so far:
dy = (2y2/x2 - 1)dx
(2y2/x2 - 1)dx - dy = 0
i used the substitution y = xv then found an integrating factor and got
dx/x - dv/(2v2 - 2v - 1) = 0
but i am stuck at this point..
i know ln(x) + C is the first part, but is there an easy way to find the second part? or am i totally off to begin with?
this is the solution:
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP50419e2227898a2f2f0000026hh297a4ebd6b3h?MSPStoreType=image/gif&s=33&w=154&h=47
i am at a loss for how they got to this point
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